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2 years ago

The approximate value of 5.1 million x 2.2 thousand is? use scientific notation

Mathematics

2 answers:

Sliva [168]2 years ago

7 0

11,220,000,000 is the answer for this question.

Nezavi [6.7K]2 years ago

7 0

The answer is 1.122*10^10

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Nady [450]

Answer:

1. 1st Quadrant

2. 3rd Quadrant

3. 2nd Quadrant

4. 2nd Quadrant

5. 4rth Quadrant

6. 2nd Quadrant

7. No they do no correspond to the same point.

Step-by-step explanation:

1st Quadrant : ( positive , positive )

2nd Quadrant : ( negative , positive )

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Pamela is making pizzas. She has 112 pieces of pepperoni. She uses 29 pieces of pepperoni on the first pizza. If she uses the sa

Svetllana [295]

Answer:

Step-by-step explanation:

Number of pieces of pepperoni available with Pamela = 112

Number of pieces of pepperoni used in one pizza = 29

Now, number of pepperoni left after using 29 in one pizza = 112 - 29 = 83

Now, Pamela makes second pizza, so again 29 pieces of pepperoni area used.

Now, number of pepperoni left after using 29 in one pizza = 83 - 29 = 54

Now, Pamela makes third pizza, so again 29 pieces of pepperoni area used.

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2 years ago

A professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their class

KIM [24]

Answer:

We conclude that seniors skip more than 2% of their classes at 0.01 level of significance.

Step-by-step explanation:

We are given that a professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their classes.

He randomly samples a group of seniors and out of 2521 classes, the group skipped 77.

Let p = percentage of seniors who skip their classes.

So, Null Hypothesis, $H_0$ : p $\leq$ 2% {means that seniors skip less than or equal to 2% of their classes}

Alternate Hypothesis, $H_A$ : p > 2% {means that seniors skip more than 2% of their classes}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of seniors who skipped their classes = $\frac{77}{2521}$

n = sample of classes = 2521

So, test statistics = $\frac{\frac{77}{2521} -0.02}{{\sqrt{\frac{\frac{77}{2521}(1-\frac{77}{2521})}{2521} } } } }$

= 3.08

The value of the test statistics is 3.08.

Now at 0.01 significance level, the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical value of z as 2.3263 < 3.08, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that seniors skip more than 2% of their classes.

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Answer:

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